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A particle has a position defined by the equations π₯ equals π‘ cubed minus five π‘ and π¦ equals three minus two π‘ squared.
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Find the velocity vector of the particle at π‘ equals two.
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Weβre told that the position of the particle is defined by a pair of parametric equations.
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In real terms, this means that, given a value of time π‘, we obtain a coordinate pair π₯π¦ for the position of the particle at that time.
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We could say that, in vector terms, the position of the particle at time π‘ is π of π‘ equals π‘ cubed minus five π‘ π plus three minus two π‘ squared π.
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This isnβt though quite what weβre looking for.
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We want to find the velocity vector of our particle when π‘ is equal to two.
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So we recall that, given a function for displacement which of course is essentially the difference in the objectβs position from one time to another, we find a function for velocity by differentiating with respect to π‘.
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This means that we can achieve a velocity vector for our particle at time π‘ by differentiating each component for the position vector with respect to π‘.
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Thatβs the derivative of π‘ cubed minus five π‘.
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And then we differentiate the vertical component as well, the derivative of three minus two π‘ squared.
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We know that to differentiate a polynomial term, we multiply the entire term by the experiment and then reduce the exposure by one.
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So π‘ cubed differentiates to three π‘ squared and negative five π‘ differentiates to negative five.
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Similarly, the derivative of three minus two π‘ squared is negative four π‘.
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Remember, the derivative of any constant is simply zero.
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Our vector function for velocity is three π‘ squared minus five π plus negative four π‘ π.
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Remember, we want a velocity vector at π‘ is equal to two.
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So weβre going to substitute π‘ equals two into our vector.
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Thatβs three times two squared minus five π plus negative four times two π, which simplifies to seven π minus eight π.